路径总和

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / 
    4   8
   /   / 
  11  13  4
 /        
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

---

题意：

给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，使得路径上所有节点值相加等于目标和。

思路：

递归地调用判断函数，根据左右子树的具体情况进行判断分类。

class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {   
        if (root == nullptr) return false;    
        return Traverse(root, 0, sum);
    }
    
    bool Traverse(TreeNode* root, int num, int target)
    {
        if (root == nullptr) return num == target;
        num += root->val;
        if (root->left != nullptr && root->right == nullptr)
        {
        return Traverse(root->left, num, target);
        }
        else if (root->right != nullptr && root->left == nullptr)
        {
        return Traverse(root->right, num, target);
        }
        else
        {
            return Traverse(root->left, num, target) ||
            Traverse(root->right, num, target);
        }
    }

};